Question

If $x+y+2=0$, then, the value of $3\left[\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}\right]$, is: ____

Answer 1

Answer: 9

We have,
$x+y+2=0\text{ (Given) }$...(1)
$\text{ Now, }3\left[\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}\right]$
$=3\left[\frac{x^2}{yz}\times \frac{x}{x}+\frac{y^2}{zx}\times \frac{y}{y}+\frac{z^2}{xy}\times \frac{z}{z}\right]$
$=3\left[\frac{x^3}{xyz}+\frac{y^3}{xyz}+\frac{z^3}{xyz}\right]$
$=3\left[\frac{x^3+y^3+z^3}{xyz}\right]$....(2)
Also,
We know that
$x^3+y^3+z^3-3xyz$
$=(x+y+z)\left(x^2+y^2+z^2-xy-yz-zx\right)$
$\Rightarrow x^3+y^3+z^3-3xyz=0\text{ (From equation (1)) }$
$\Rightarrow x^3+y^3+z^3=3xyz$...(3)
Putting value of equation (3) in equation (2)
$3\left[\frac{3xyz}{xyz}\right]=3\times 3=9$